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Line L has an undefined slope. Line M is perpendicular to line L. Which of the following could be the equation of line M?

Marat540 [252]

"Line L has an undefined slope" so line L is any vertical line.

Anything perpendicular to this vertical line will be horizontal. So line M is horizontal.

Horizontal lines are of the form y = k for some fixed number k. The only thing that fits is choice B) y = 7 where k = 7 in this case.

Choice A graphs out a diagonal line, choice C graphs a vertical line, while choice D graphs out a hyperbola. These facts are why we can eliminate these non-answers.

Answer: Choice B) y = 7

6 0

3 years ago

Que is on pic.i can't able to type in text.

ad-work [718]

It's not difficult to compute the values of $A$ and $B$ directly:

$A=\displaystyle\int_1^{\sin\theta}\frac{\mathrm dt}{1+t^2}=\tan^{-1}t\bigg|_{t=1}^{t=\sin\theta}$
$A=\tan^{-1}(\sin\theta)-\dfrac\pi4$

$B=\displaystyle\int_1^{\csc\theta}\frac{\mathrm dt}{t(1+t^2)}=\int_1^{\csc\theta}\left(\frac1t-\frac t{1+t^2}\right)\,\mathrm dt$
$B=\left(\ln|t|-\dfrac12\ln|1+t^2|\right)\bigg|_{t=1}^{t=\csc\theta}$
$B=\ln\left|\dfrac{\csc\theta}{\sqrt{1+\csc^2\theta}}\right|+\dfrac12\ln2$

Let's assume $0$ , so that $|\csc\theta|=\csc\theta$ .

Now,

$\Delta=\begin{vmatrix}A&A^2&B\\e^{A+B}&B^2&-1\\1&A^2+B^2&-1\end{vmatrix}$
$\Delta=A\begin{vmatrix}B^2&-1\\A^2+B^2&-1\end{vmatrix}-e^{A+B}\begin{vmatrix}A^2&B\\A^2+B^2&-1\end{vmatrix}+\begin{vmatrix}A^2&B\\B^2&-1\end{vmatrix}$
$\Delta=A(-B^2+A^2+B^2)-e^{A+B}(-A^2-A^2B-B^3)+(-A^2-B^3)$
$\Delta=A^3-A^2-B^3+e^{A+B}(A^2+A^2B+B^3)$

There doesn't seem to be anything interesting about this result... But all that's left to do is plug in $A$ and $B$ .

3 0

2 years ago

Ok since no one answered my previous question i don't wanna put to waste so f r ee p oi nts yeah

Zolol [24]

Answer:

yea its jaiashhsjajajsjs

6 0

3 years ago

Read 2 more answers

Help...please?? Thanks guys

vfiekz [6]

Answer:

x=20

Step-by-step explanation:

The Pythagorean theorem is

a^2 +b^2 = c^2

where a and b are the legs and c is the hypotenuse.

The legs are x and 15 and the hypotenuse is 25

x^2 + 15^2 = 25^2

x^2 +225 = 625

Subtract 225 from each side

x^2 +225-225 = 625-225

x^2 = 400

Take the square root of each side

sqrt(x^2) = sqrt(400)

x = 20

4 0

3 years ago

Read 2 more answers

If u, v, and w are nonzero vectors in r 2 , is w a linear combination of u and v?

Tju [1.3M]

Not necessarily. $\mathbf u$ and $\mathbf v$ may be linearly dependent, so that their span forms a subspace of $\mathbb R^2$ that does not contain every vector in $\mathbb R^2$ .

For example, we could have $\mathbf u=(0,1)$ and $\mathbf v=(0,-1)$ . Any vector $\mathbf w$ of the form $(r,0)$ , where $r\neq0$ , is impossible to obtain as a linear combination of these $\mathbf u$ and $\mathbf v$ , since

$c_1\mathbf u+c_2\mathbf v=(0,c_1)+(0,-c_2)=(0,c_1-c_2)\neq(r,0)$

unless $r=0$ and $c_1=c_2$ .

8 0

3 years ago